Arithmetic Progressions Test - 15

Hint: In this question, we need to find the sum of the ${25}^{th}$ and ${30}^{th}$ term of the given sequence. So, first, we will calculate the $n^{th}$ term for both numbers of terms. After that, we will be adding them.

Step by step solution:

The general or nth term of an AP is given as

$T=a+(n-1) d$

where

a= first term

d= common difference

Substitution the given values

So, the 25 ^th term of the given sequence will be:

$a = 3$

$n = 25$

$d = 3$

$T=a+(n-1) d$

$T = 3 + (25 - 1) \times 3$

$T = 3 + 75 -3 = 75$

Similarly, we will take out the 30th term of it.

$n = 30$

$T = 3 + (30 - 1) \times 3$

$T = 3 + 90 - 3 = 90$

the sum of 25th and 30th term will be=

$75 + 90 = 165$

Note: While solving this question, the student must keep in mind that, they will sum of ${25}^{th}$ and ${30}^{th}$ terms after finding ${25}^{th}$ and ${30}^{th}$ term.

Therefore, option B is correct.

Hint: In this question, we need to find the selling price of the machine. After that we will calculate the total profit. Similarly, we can calculating the cost price of the machine, including repair cost.

Step by step Solution:

The total selling price of $30$ machines 

$= 30 \times 300 = Rs. 9000$

Total profit $= 150 \times 30 = Rs. 4500$

Total cost price of $30$ machines (including repair cost) $= 9000 - 4500 = Rs. 4500$

Amount spent on repairs

$= 4500 – 2000 = Rs. 2500$

Note: While solving this question students note that, A man bought $30$ defective machines for $Rs. 2000$ but he repaired the machine and sold it at $Rs. 300$ per machine. So, when we will be calculating the cost price of $30$ machines also include repairing cost.

Hence, the correct option is (C).

Hint: In this question, we need to find the number of terms. We will calculate the $n^{th}$ number of terms by using the general formula of AP.

Step by step solution:

Given sequence is $1, 7, 13,..., 235$

The first term $a = 1$ and 

the common difference $d = 7 - 1 = 6$

the last term be $235$

Let the last term be the nth term

We know that the nth term of the arithmetic progression is given by $a + (n - 1) d$

Therefore, $a + (n - 1) d = 235$

$= 1 + (n - 1) \times 6 = 235$

$= 1 + 6n - 6 = 235$

$= 6n - 5 = 235$

$= 6n = 240$

$= n = 40$

${40}^{th}$ term of the given sequence is $235$.

Note: While solving this question, students not that, in the given sequence the value of $235$ used as the last term.

Hence, the correct option is (D).

Hint: It can be seen from the given question, that the incomes of Subha Rao increase every year by $Rs. 200$ and hence, form an AP.

Therefore, after $1995$, the salaries for each year are:

$5000, 5200, 5400, ...$

Step by step solution:

Subba Rao's starting salary $=Rs. 5000$

Annual increment $=Rs. 200$

Let $n$ denote the number of years.

First term $=a=Rs. 5000$

Common difference $=d=Rs. 200$

$a_n=Rs. 7000$

$\therefore 5000+\left(n-1\right)200=7000 \left[\because a_n=a+\left(n-1\right)d\right]$

$\Rightarrow \left(n-1\right)200=2000$

$\Rightarrow n-1=\frac{2000}{200}=10$

$\Rightarrow n=10+1=11$

Hence, in the ${11}^{th}$ year, his salary will become $Rs. 7000$.

Now, in the case of the year, the sequence is $1995, 1996, 1997, 1998,...$

Let $a_n$ a denote the required year.

$a_n=1995+\left(11-1\right)1=1995+10=2005$

Hence, in the year $2005$, Subba Rao's salary becomes $Rs. 7000$.

Note: While solving this question, students must note that the value of $a_n$ is $Rs. 7000$ and first term (a) is $Rs. 5000$

Hence, the correct option is (D).

Hint: In this question, we need to find the ${80}^{th}$ term of a given sequence by using the general formula, 

$T=a+\left(n-1\right)d$

In this sequence the value of $a=28$ and $d=a_2-a_1=6$

The general or nth term of an AP is given as

$T=a+(n-1) d$

where

a= first term

d= common difference

$n = 80$

Substitution of the given values

$T = 28 + (80 - 1) \times 6$

$T = 28 + 480 - 6 = 502$

Note: While solving this question students must keep in mind that, they have to find the numbers of terms, not the n^th term is already given in this question.

Therefore, option A is correct.

Hint: The odd numbers between $0$ and $50$ are $1, 3, 5, 7, 9,... 49$.

Therefore, we can see that these odd numbers are in the form of AP.

Step by step solution:

We know that the first term of an odd number, $a=1$

Common difference, $d=2$

Last term, $l=49$

By the formula of the last term, we know

$l=a+\left(n-1\right)d$

$49=1+\left(n-1\right)2$

$48=2\left(n-1\right)$

$n-1=24$

$n=25=$ Number of terms

By the formula of the sum of $n^{th}$ term, we know

$S_n=\frac{n}{2}\left(a+l\right)$

$S_{25}=\frac{25}{2}\left(1+49\right)$

$=25\frac{\left(50\right)}{2}$

$=\left(25\right)\left(25\right)$

$=625$

Note: While solving this question, the students should keep in mind that, the last terms of an odd number from $0$ to $50$ is $49$, not $50$.

Hence, the correct option is (C).

Hint: In this question, Raven traveled $5 km$ on the first day. He kept on increasing his traveling distance by $2 km$ each day. So, on the second day, his traveling distance $5+2=7$ km. Similarly, from the third day to the ${12}^{th}$ days, we will sum the traveling distance, that traveled by Raven.

Step by step solution:

For the first day, Raven traveled $= 5 km$ (given)

For the second day, Raven traveled $2 km$ more $= 5 + 2$

$= 7 km$

According to the question every day he walks $2 km$ more.

So, $1^{st}$ day $+ 2 km + 2 km$........ ${12}^{th}$ day

$5+2+2+2+2+2+2+2+2+2+2+2= 27 km$

Note: While solving this question, the students must keep in mind that, Raven had traveled $5 km$ only on the first day. After that, he kept on increasing his traveling distance by $2 km$ each day.

Hence, the correct option is (C).

Hint: The positive integers that are divisible by $6$ are $6, 12, 18, 24...$

We can see here, that this series forms an AP. Whose first term is $6$ and the common difference is $6$.

Step by step solution:

First-term of positive integers divisible by $6$,

$d=6$

$S_{40}=$?

By the formula of the sum of $n$ terms, we know

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Therefore, putting $n=40$, we get

$S_{40}=\frac{40}{2}\left[2\left(6\right)+\left(40-1\right)6\right]$

$=20\left[12+\left(39\right)\left(6\right)\right]$

$=20\left(12+234\right)$

$=20\times 246$

$=4920$

Note:  While solving this question, students must keep in mind that, Integers are like whole numbers, but they also include negative numbers.

Hence, the correct option is (C).

Hint: In this question, we need to find the ${29}^{th}$ term. So, we can use the third term and last term to calculating the ${29}^{th}$ term.

Step by step solution:

Let $a$ and $d$ be the first term and common difference of the given AP.

Given that, $a_3=12$

$\Rightarrow a+\left(3-1\right)d=12 \left[\therefore a_n=a+\left(n-1\right)d\right]$

$\Rightarrow a+2d=12$...(i)

Last term $=T_{50}=106$

$\Rightarrow a+\left(50-1\right)d=106$

$\Rightarrow a+49d=106$...(ii)

Subtracting (i) from (ii), we have

$\Rightarrow 47d=94$

$\Rightarrow d=\frac{94}{47}=2$

Putting the value of $d$ in (i), we have

$a+2\left(2\right)=12$

$\Rightarrow a+4=12$

$\Rightarrow a=8$

Now, $a_{29}=a+\left(29-1\right)d$

$=8+28\left(2\right)=8+56=64$

Note: While solving this question, students must keep in mind that, the AP consists of $50$ terms. So, we can use the value of $n=50$, for the calculating last term.

Hence, the correct option is (C).

Hint: In this question, we need to find, the sum of the first $22$ terms. So, we will use the formula, sum of $n$ terms $\left(S_n\right)=\frac{n}{2}\left(a+a_n\right)$.

Step by step solution:

Given, 

The common difference, $d=7$

${22}^{nd}$ term $a_{22}=149$

Sum of first $22$ terms, $S_{22}=$?

By the formula of $n^{th}$ term,

$a_n=a+\left(n-1\right)d$

$a_{22}=a+\left(22-1\right)d$

$149=a+147$

$a=2=$ First term

Sum of $n$ terms,

$S_n=\frac{n}{2}\left(a+a_n\right)$

$S_{22}=\frac{22}{2}\left(2+149\right)$

$=11\times 151$

$=1661$

Note: While solving this question, the student must keep in mind that first, we will find the value of $a$ after that put this value in $\frac{n}{2}\left(a+a_n\right)$ formula.

Hence, the correct option is (A).

Hint: In this question, the difference between them is known by looking at the given number sequence.

Step by step solution:

We know that $d=a_2-a_1=a_{3-}{a}_2=a_4-a_3$

$a_1=60, a_2=50, a_3=40$ and $a_4=30$

So, $d=50-60=40-50=30-40=-10$

Thus, this given sequence subtracting $10$ from each terms.

Note: While solving this question, students must keep in mind that, Always subtract the second term from the first and the third term from the second term.

Hence, the correct option is (B).

Hint: In this question, David plucked $50$ apples from a tree on the first day. From the second to the sixth day, he kept on plucking $15$ apples every day. So after $6$ days, to calculate the number of apples, we have to add the apples that have been broken in all the days.

Step by step solution:

On the first day, David plucked apples $= 50$ (given)

On the second day, he plucked $15$ apples.

Similarly, on the third day, he plucked $15$ more apples, and so on till $6$ days are completed.

So, it will be

$1^{st} day + 2^{nd} day + 3^{rd} day +.......... 6^{th} day$

$50 + 15 + 15 + 15 + 15 + 15 = 125$ apples

Note: While solving this question, students must keep in mind that, David has broken $50$ apples on the first day while in the remaining $5$ days he has broken $15$ apples each day.

Hence, the correct option is (B).

Hint: In this question, we need to find the number of terms. So, we can use the general formula for the $n^{th}$ term, 

$a_n=a+\left(n-1\right)d$

Step by step solution:

Given, AP is $7, 13, 19,..., 205$

$a=7$

$d=13-7=6, a_n=205$

Using formula, $a_n=a+\left(n-1\right)d$

$\Rightarrow 205=7+\left(n-1\right)6$

$\Rightarrow \left(n-1\right)6=205-7$

$\Rightarrow \left(n-1\right)=\frac{198}{6}=33$

$\Rightarrow n=33+1=34$

Hence, ${34}^{th}$ term of the AP is $205$.

Note: While solving this question, students must keep in mind that, the value of n is unknown and the $a_n$ ($n^{th}$ term in the sequence) is given.

Hence, the correct option is (D).

Hint: In this question, all terms in the given sequence are negative. We will solve this type of question using the general formula of the $n^{th}$ term.

Step by step solution:

The general or $n^{th}$ term of an AP is given as

$T=a+(n-1) d$

where

$a=$ first term

$d=$ common difference

Substitution the given values

$T = (-9) + (n-1) \times -4$

$T = (-9) - 4n + 4$

$T = -4n - 5$

Note: Students don't be afraid to see all the terms in question negatively. Students should solve this question like a normal question.

Hence, the correct option is (B).

Hint: In this question, the ${17}^{th}$ term of an AP exceeds its ${10}^{th}$ term by $7$.

According to question, $a_{17}-a_{10}=7$

Step by step solution:

Let $a$ and $d$ be the first term and common difference of the given AP.

$a_{17}=a+\left(17-1\right)d=a+16d$

$a_{10}=a+\left(10-1\right)d=a+9d$

As per condition, $a_{17}-a_{10}=7$

$\Rightarrow \left(a+16d\right)-\left(a+9d\right)=7$

$\Rightarrow 7d=7$

$\Rightarrow d=\frac{7}{7}=1$

Hence, common difference is $1$.

Note: While solving this question, students must keep in mind that, when calculating common difference between the ${17}^{th}$ term and $10$ terms, subtract the ${17}^{th}$ term by $10$ terms.

Hence, the correct option is (B).

Hint: In this question, we need to calculate the value of $n$ by using the general formula of $n^{th}$ terms.

Step by step solution:

Let $a=$ Amount saved in the first-week $=Rs. 5$

$d=$ increased in saving every week $=Rs 1.75$

$\therefore 5+\left(n-1\right)1.75=20.75$

$\left[\because a_n=a+\left(n-1\right)d\right]$

$\Rightarrow \left(n-1\right)1.75=20.75-5$

$\Rightarrow \left(n-1\right)=\frac{1575}{100}\times \frac{100}{175}$

$\Rightarrow n-1=9$

$\Rightarrow n=9+1=10$

So, in ${10}^{th}$ week, Ramkali's saving becomes $Rs 20.75$.

Hence, the correct option is (A).

Hint: In this question, we need to calculate the value of $t_1$, and $t_3$ by using the general formula of $n^{th}$ terms.

Step by step solution:

Given:

$t_2=13$ and $t_4=3$

Then, $t_2=a+\left(2-1\right)d$

$\Rightarrow 13=a+d$...(i)

$t_4=a+\left(4-1\right)d$

$\Rightarrow 3=a+3d$...(ii)

Subtracting equation (i) from equation (ii), we get

$d=-5$

Putting $d=-5$ in equation (i), we get

$a=13+5=18$

$t_3=a+\left(3-1\right)d$

$=18+2\times \left(-5\right)$

$=18-10$

$=8$

Hence, the complete sequence is $18, 13, 8, 3$.

Note: While solving this question, students must keep in mind that, The value of $d$ can be positive or negative.

Hence, the correct option is (C).

Hint: The multiples of $8$ are $8, 16, 24, 32,...$

The series is in the form of AP, having first as $8$ and a common difference as $8$.

Step by step solution:

We know that, the first term of multiples of $8$, $a=8$

$d=8$

$S_{15}=$?

By the formula of the sum of $n^{th}$ term, we know

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

$S_{15}=\frac{15}{2}\left[2\left(8\right)+\left(15-1\right)8\right]$

$=\frac{15}{2}\left[16+112\right]$

$=15\frac{\left(128\right)}{2}$

$=15\times 64$

$=960$

Note: While solving this question, the student must keep in mind that, the meaning of multiplication of 8 is the same as skip counting by a number 8 times Or you can also think of it as adding 8 equal groups.

Hence, the correct option is (B).

Hint: In this question, we need to find the $n^{th}$ terms, so we will calculate the $n^{th}$ terms by using the general formula.

Step by step solution:

The general or nth term of an AP is given as

$T=a+(n-1) d$

where

a= first term

d= common difference

Substitution the given values

$T = 4 + (n -1) \times 6$

$T = 4 + 6n - 6 = 6n -2$

Note: In the given question, the $n$ term and number of terms both are not given.

Therefore, option D is correct.

Hint: In this question, the given sequence is $4, 6, 8, 10...n^{th}$. we can find the $n^{th}$ term of this given sequence, we will use to general formula $\left[T=a+\left(n-1\right)d\right]$ of an Arithmetic progression.

Step by step solution:

The general or $n^{th}$ term of an AP is given as

$T=a+(n-1) d$

where

$a=$ first term

$d=$ common difference

Substitution the given values

$T = 4 + (n-1) \times 2$

$T = 4 + 2n - 2$

$T = 2n + 2$

Note: In this given sequence, the value of $d$ is $2$. 

$\therefore d=a_2-a_1=a_3-a_2...$

Hence, the correct option is (C).

Hint: In the question, we know that the first terms of a given sequence. So, now we need to find the common difference,

Common difference, $d=a_2-a_1=a_3-a_2$

Step by step solution:

Given AP is $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3},...$

Here, $a_1=\frac{1}{3}, a_2=\frac{5}{3}, a_3=\frac{9}{3}, a_4=\frac{13}{3}$

First term $=a_1=\frac{1}{3}$

$d=a_2-a_1=\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}$

$=a_3-a_2=\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}$

$=a_4-a_3=\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}$

Note: While solving this question, students don't be confused by looking at the question in friction form. We can solve this question in the normal method.

Hence, the correct option is (A).

Hint: In this question, we need to find the $9^{th}$ term of a given sequence. So, we can use the general formula for $n^{th}$ term $T=a+\left(n-1\right)d$.

Step by step solution:

The general or $n^{th}$ term of an AP is given as

$T=a+(n-1) d$

where

$a=$ first term

$d=$ common difference

$n= 9^{th}$ (given)

Substitution the given values

$T = 9 + (9-1) \times 9$

$T = 9 + 8 \times 9$

$T = 9 + 72 = 81$

Note: While solving this question, students must keep in mind that, the value of the first term and $n^{th}$ term, both are the same.

Hence, the correct option is (D).

Hint: In this question, we need to find the sum of the first $15$ natural numbers. We will calculate by using the following formula:

$s=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Step by step solution:

AP $=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$

Given, $a=1, d=2-1=1$ and $a_n=15$

Now, by the formula, we know,

$s=\frac{n}{2}\left[2a+\left(n-1\right)\times d\right]=\frac{15}{2}\left[2\times 1+\left(15-1\right)\times 1\right]$

$s=\frac{15}{2}\left[2+14\right]=\frac{15}{2}\left[16\right]=15\times 8$

$s=120$

Hence, the sum of the first $15$ natural numbers is $120$.

Note: While solving this question, students should know about natural numbers.

Hence, the correct option is (C).

Hint: In the question, we need to find the ${30}^{th}$ term of the given AP. So, we will use the general formula $\left[a_n=a_1+\left(n-1\right)d\right]$ for finding the ${30}^{th}$ term.

Step by step solution:

Given, AP is $10, 7, 4,...$

$a=10$

$\because d=7-10=4-7=-3$

$a_n=a+\left(n-1\right)d$

Now, $a_{30}=10+\left(30-1\right)\left(-3\right)=10-87=-77$

Note: While solving this question, the student must keep in mind that, the value of $n$ is $30$.

Hence, the correct option is (C).

Hint: In this question, we need to find the sum of the square of the ${20}^{th}$ and ${10}^{th}$ terms of the given sequence. So, first, we will calculate $n^{th}$ terms for both numbers of terms, after that squaring both the numbers and adding them.

Step by step solution:

The general or nth term of an AP is given as

$T=a+(n-1) d$

where

a= first term

d= common difference

Substitution the given values

So, the 20 ^th term of the given sequence will be:

$a = 60$

$n = 20$

$d = -4$

$T=a+(n-1) d$

$T = 60 + (20 - 1) \times (-4)$

$T = 60 - 80 + 4$

$T = 64 - 80 = -16$

Similarly, we will take out the ${10}^{th}$ term of it.

$n = 10$

$T = 60 + (10 - 1) \times (-4)$

$T = 60 - 40 + 4 = 24$

square of 20th and 10th term will be = ${\left(-16\right)}^2+{\left(24\right)}^2 = 256 + 576$ = $832$

Note: While solving this question, the student must keep in mind that, they will sum of square of ${20}^{th}$ and ${10}^{th}$ terms after finding ${20}^{th}$ and ${10}^{th}$ terms.

Hence, the correct option is (B).